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GMAT Prep - Geometry

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by akram » Fri Aug 24, 2007 4:09 pm
:arrow:
Last edited by akram on Fri Aug 24, 2007 4:13 pm, edited 1 time in total.
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Re: GMAT Prep - Geometry

by akram » Fri Aug 24, 2007 4:12 pm
tutonaranjo wrote:No luck solving this. Please help.
Image
Options are:
a) 1/2
b) 1
c) sqrt 2
d) sqrt 3
e) sqrt 2 / 2
B is correct answer.

(s,t) will be (1, sqrt 3), radius is 2 .
Angle O is 60 degress on left and 30 degrees on right since sin 360=sqrt 3/2
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Help with explanation

by tutonaranjo » Fri Aug 24, 2007 5:01 pm
Got it.... 30:60:90 triangles... should've spotted that when seing 1, sqrt 3
thanks
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by koushik » Tue Aug 28, 2007 6:35 pm
In the fig. side OP = OQ, then shouldn't angle OPQ = OQP = 45 degree ?
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Re: GMAT Prep - Geometry

by Jameschan168 » Wed Aug 29, 2007 1:09 pm
I think another way to look at it is this.

The 2 lines are perpendicular to each other, so their slopes are the negative inverse to each other. Also since they are both the radius of the semi-circle, they are both equidistant to the origin.

Hence, s,t has the coordinate of (1,sq 3).
akram wrote:
tutonaranjo wrote:No luck solving this. Please help.
Image
Options are:
a) 1/2
b) 1
c) sqrt 2
d) sqrt 3
e) sqrt 2 / 2
B is correct answer.

(s,t) will be (1, sqrt 3), radius is 2 .
Angle O is 60 degress on left and 30 degrees on right since sin 360=sqrt 3/2
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Re: GMAT Prep - Geometry

by kadishmj » Wed Sep 05, 2007 12:45 pm
akram wrote:
tutonaranjo wrote:No luck solving this. Please help.
Image
Options are:
a) 1/2
b) 1
c) sqrt 2
d) sqrt 3
e) sqrt 2 / 2
B is correct answer.

(s,t) will be (1, sqrt 3), radius is 2 .
Angle O is 60 degress on left and 30 degrees on right since sin 360=sqrt 3/2
B is correct, but akram's reasoning is not. I have no idea where he comes up with sin 360 = sqrt 3/2

It is a 30-60-90 triangle, and Angle O is 30 degrees on the left and 60 on the right (not what akram said). This may clear things up:

https://www.themathpage.com/aTrig/30-60-90-triangle.htm
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by ri2007 » Wed Sep 05, 2007 1:15 pm
tutonaranjo can you pls confirm the right answer, I keep getting it as sq root of 3. I can explain how if this is correct. If this is wrong I need to understand where i went wrong.

thanks
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by krishnamurthyu » Thu Sep 06, 2007 5:34 am
Perpendicular Property:
If the angle b/w 2 points P (x1,y1) and Q (x2,y2) is 90 i.e right angle then
then the product of their slopes is -1 i.e y1/x1 * y2/x2 = -1 .
y2/x2 = -1 x1/y1;
= -1 -RT(3)/1
y2/x2 =RT(3)/1 =====> Q(1,RT(3))

Easy to Remember:
if Point P (a,b) then it's Perpendicular point is (-b,a) :
in this case : P (-RT(3),1 ) , Q(s,t) (1,RT(3) )
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by ri2007 » Thu Sep 06, 2007 5:59 am
thanks a lot krishnamurthyu for your reply and explaination
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by bkabbani » Sun Sep 09, 2007 9:32 am
The rule that krishnamurthyu mentioned is correct but the way to remember it could be misleading.

In the case of this problem if point P coordinates are (a,b), then point Q coordinates are (b,-a) NOT (-b,a).

It really depends on the quadrant where point Q lies. If it was in QIII then the coordinates would have been (-b,a).
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Diagram is misleading

by Kansonne » Fri Dec 28, 2007 3:22 pm
Hi,

I found that while the diagrams are generally helpful and accurate on the GMAT, in this particular case, the diagram was misleading.

I was stuck on D (root 3) for an answer until I saw from one of the previous posts, using the 30-60-90 rule, that the "right half" of angle POQ MUST be be 30 degrees.

From this I quickly realized that while the diagram make it look like the right angle at POQ is bisected into two 45 degree angles, that is NOT the case. I think root 3 is the trap answer for those of us who mistakenly thought PQ is parallel to the x-axis, and that t equals 1. Really, the only thing to be had from P is that the radius of the circle (side OP) is 2.

Using the 30-60-90 rule, we get s = 1.

Tricky.
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by sibbineni » Sun Jan 06, 2008 1:34 pm
i think koushik is absolutely write


In the fig. side OP = OQ=2...

since 2 sides are equal it is an isoceles triangle
then angle OPQ = OQP = 45 degree ...

since we know length of 2 sides then length of PQ=?

by pythagaros theorem

PQ^2=OP^2+OQ^2
=4+4
=8
PQ=2sqrt(2)

there fore length of PQ=2sqrt(2)

Formula:
Distance between 2 points (x1,y1) and (x2,y2) is sqrt (x2-x1)^2+(y2-y1)^2....

then calculate the distance between O and Q

since O is origin theno(0,0)
Q(s,t)

after calculating the distance then
s^2+t^2=4----(1)

then the distance of P and Q

s^2+t^2+t+4+s sqrt(3)=8----(2)

substiting the 1 value in 2 we have

4+t+4+s sqrt(3)=8
t=- s sqrt(3) ---- (3)

substiting the value of 3 in 1 we have
s^2+3 s^2=4
4s^2=4
s=1

and substitute the value s=1 in eq 1 we have

t=sqrt(3)
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by lunarpower » Wed May 13, 2009 2:41 am
here's an awesome way to approach this.

see that 90 degree angle there? ok. that means that this question is really asking where the point (-rad3, 1) would go if the paper were rotated clockwise by 90 degrees.

so...

draw that point on your paper, and then physically rotate the paper by 90 degrees.

originally the x coordinate was -√3 (to the left). when you rotate the paper, this is now upward, so it's positive √3 in the y direction.
originally the y coordinate was positive 1 (upward). when you rotate the paper, this is now to the right, so it becomes positive 1 in the x direction.

ergo, new coordinates = (1, √3)

sweet
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by vineetbatra » Thu Jun 25, 2009 7:32 pm
krishnamurthyu wrote:Perpendicular Property:
If the angle b/w 2 points P (x1,y1) and Q (x2,y2) is 90 i.e right angle then
then the product of their slopes is -1 i.e y1/x1 * y2/x2 = -1 .
y2/x2 = -1 x1/y1;
= -1 -RT(3)/1
y2/x2 =RT(3)/1 =====> Q(1,RT(3))

Easy to Remember:
if Point P (a,b) then it's Perpendicular point is (-b,a) :
in this case : P (-RT(3),1 ) , Q(s,t) (1,RT(3) )
Krishanmurhyu, the slope of a line is Y2-Y1/x2-x1 , for a given line we have only one coordinates (i.e. P or Q), unless we are assuiming that O is (0,0).

Please explain?

Thanks,

Vineet
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